P6601

标签：P6601 2p sum choose frac 2n 2i

我们发现每一时刻的小球位置只可能有两种，这和它瞬移的次数有关。在每个时刻内，都有两种可能的方案。对于每个时刻瞬移次数为奇数的概率就是\(\sum_{i=0}^{t} {n \choose i} p^{i}*(1-p)^{t-i}[i\%2==1]\),偶数就是\(\sum_{i=0}^{t} {n \choose i} p^{i}*(1-p)^{t-i}[i\%2==0]\)
根据概率的定义和古典概型(每种方案选到的概率相等)，是总概率除以总方案数。那么答案就是\(\frac{1}{2n(t+1)} \sum_{i=0}^{t} even_{i}^{2}+odd_{i}^{2}\)考虑将mod 2拆掉\(\frac{1-(-1)^{i}}{2}\)那么答案就是\(\frac{1}{2n(t+1)} \sum_{i=0}^{t} （\frac{1-(-1)^{i}}{2}*\sum_{i=0}^{t} {n \choose i} p^{i}*(1-p)^{t-i}）^{2}+（\frac{1+(-1)^{i}}{2}\sum_{i=0}^{t} {n \choose i} p^{i}*(1-p)^{t-i}）^{2}\)

\[\sum_{i=0}^{t} {n \choose i} p^{i}*(1-p)^{t-i}[i\%2==1] \]

\[\sum_{i=0}^{t} {n \choose i} p^{i}*(1-p)^{t-i}*\frac{1-(-1)^{i}}{2} \]

\[\frac{1}{2}\sum_{i=0}^{t} {n \choose i} p^{i}*(1-p)^{t-i} -\frac{1}{2}\sum_{i=0}^{t} {n \choose i} (-p)^{i}*(1-p)^{t-i} \]

\[\frac{1}{2}-\frac{1}{2}(1-2p)^{t} \]

这就是奇数的情况

\[\frac{1}{2}+\frac{1}{2}(1-2p)^{t} \]

\[\frac{1}{2n(t+1)} \sum_{i=0}^{t} even_{i}^{2}+odd_{i}^{2} \]

\[\frac{1}{2n(t+1)} \sum_{i=0}^{t} (\frac{1}{2}-\frac{1}{2}(1-2p)^{t})^{2}+(\frac{1}{2}+\frac{1}{2}(1-2p)^{t})^{2} \]

设$$x=\frac{1}{2},y=\frac{1}{2}(1-2p)^{t}$$
原式

\[=\frac{1}{2n(t+1)} \sum_{i=1}^{t+1} (x-y)^{2}+(x+y)^{2} \]

\[=\frac{1}{n(t+1)} \sum_{i=1}^{t+1} x^{2}+y^{2} \]

\[=\frac{1}{4n(t+1)} \sum_{i=1}^{t+1} 1+(1-2p)^{2i} \]

\[=\frac{1}{4n(t+1)} (t+1+\sum_{i=1}^{t+1} (1-2p)^{2i}) \]

等比数列求和
设 $$S=\sum_{i=1}^{t+1} (1-2p)^{2i}$$

\[(1-2p)^{2}*S=\sum_{i=1}^{t+2} (1-2p)^{2i} \]

\[S*((1-2p)^{2}-1)=(1-2p)^{2t+4}-(1-2p)^{2} \]

\[S=\frac{(1-2p)^{2t+4}-(1-2p)^{2}}{(1-2p)^{2}-1} \]

参考nacly_fish的题解

标签：P6601,2p,sum,choose,frac,2n,2i
From： https://www.cnblogs.com/wuhupai/p/18341911